If \( f(x)=\begin{cases}\displaystyle\frac{1-\cos 2kx}{x^2}, & \text{if}\ x\ne 0\\8, & \text{if}\ x=0\end{cases} \) is a continuous function at \( x=0 \), then the value of \( k \) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \pm 2 \)
For continuity at x = 0, the limit of f(x) as x → 0 must equal f(0) = 8. By \( 1-\cos 2\theta=2\sin^2\theta \):
\( \displaystyle \lim_{x\to 0}\frac{1-\cos 2kx}{x^2}=\lim_{x\to 0}\frac{2\sin^2 kx}{x^2} \)
\( \displaystyle =2k^2\lim_{x\to 0}\left(\frac{\sin kx}{kx}\right)^2 \)
\( \displaystyle =2k^2 \)
\( 2k^2=8 \)
\( k^2=4 \)
\( k=\pm 2 \)
Why the other options are incorrect
Option (2) \( \pm 2\sqrt{2} \) — Incorrect. \( k=\pm 2\sqrt{2} \) comes from \( k^2=8 \), which forgets the factor 2 in \( 1-\cos 2kx=2\sin^2 kx \).
Option (3) \( \pm 1 \) — Incorrect. With \( k=\pm 1 \) the limit is \( 2(1)^2=2 \), not 8.
Option (4) \( \pm 4 \) — Incorrect. With \( k=\pm 4 \) the limit is \( 2(4)^2=32 \), not 8.